Question

Ap Calculus B.C. This involves improper integrals. I need help so give me a detailed explanation with steps.

Answer 1

`\int ^(-1)_(-\infty)e^(5t)dt`

The first thing we are going to do is a substitution:

`\begin{gathered} u=5t \\ du=5dt\to dt=(1)/(5)du \end{gathered}`

We replace the substitution in the integral and solve the integral

`\begin{gathered} \int ^(-1)_(-\infty)e^u\cdot((1)/(5)\cdot du) \\ (1)/(5)\int ^(-1)_(-\infty)e^u\cdot\mathrm{d}u \end{gathered}`

We know that the integral of an exponential is the same exponential, so:

`(1)/(5)\int ^(-1)_(\infty)e^u\cdot\mathrm{d}u=(1)/(5)e^u`

To evaluate the answer we must recover the substitution made from the beginning:

`(1)/(5)e^u=(1)/(5)e^(5t)`

Now, we evaluate in the interval from negative infinity to negative 1

`\begin{gathered} \int ^(-1)_(-\infty)e^(5t)dt=(1)/(5)e^(5t) \\ \int ^(-1)_(-\infty)e^(5t)dt=(1)/(5)e^(5(-1))-(1)/(5)e^(5(-\infty)) \\ \int ^(-1)_(-\infty)e^(5t)dt=(1)/(5)e^(-5)-0 \end{gathered}`
`\int ^(-1)_(-\infty)e^(5t)dt=(1)/(5)e^(-5)=0.00135`

This integral is convergent